Question

How many grams of dry NH4Cl need to be added to 1.60 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.81? Kb for ammonia is 1.8×10−5.

Answer 1

  NH4Cl ----------> NH4(+) + Cl(-)
NH3 + H2O <----------> NH4(+) + OH(-)
The Kb is small so the it is possible to say that all the NH4(+) is released from NH4Cl and the concentration of NH3 will not change

pOH = 14-pH

pOH = 14- 8.81

pOH = 5.19

from this concentration of OH-

[OH-] = 10^-pOH

[OH-] = 10^-5.19

[OH-] = 6.46 x 10^-6

Kb = [OH-][NH4+] / [NH3]

1.8×10⁻⁵ = [6.46 x 10^-6][NH4+] / 0.8

[NH4+]= 1.44 x 10^-5 / 6.46 x 10^-6

[NH4+] = 2.23M

from molarity we can calculate the molarity

2.23 = = (weight of NH4Cl / molar mass of NH4Cl) x 1/V in liters

weight of NH4 Cl = 2.33 M x 1.6 L x 53.5

= 199.45 grams